Consider the functional

$$I[y]=\int_{-\infty}^{\infty}\left(\frac{1}{2} y^{\prime 2}+\frac{1}{2} U(y)^{2}\right) d x$$

where $y(x)$ is subject to boundary conditions $y(x) \rightarrow a_{\pm}$as $x \rightarrow \pm \infty$ with $U\left(a_{\pm}\right)=0$. [You may assume the integral converges.]

(a) Find expressions for the first-order and second-order variations $\delta I$ and $\delta^{2} I$ resulting from a variation $\delta y$ that respects the boundary conditions.

(b) If $a_{\pm}=a$, show that $I[y]=0$ if and only if $y(x)=a$ for all $x$. Explain briefly how this is consistent with your results for $\delta I$ and $\delta^{2} I$ in part (a).

(c) Now suppose that $U(y)=c^{2}-y^{2}$ with $a_{\pm}=\pm c(c>0)$. By considering an integral of $U(y) y^{\prime}$, show that

$$I[y] \geqslant \frac{4 c^{3}}{3},$$

with equality if and only if $y$ satisfies a first-order differential equation. Deduce that global minima of $I[y]$ with the specified boundary conditions occur precisely for

$$y(x)=c \tanh \left\{c\left(x-x_{0}\right)\right\}$$

where $x_{0}$ is a constant. How is the first-order differential equation that appears in this case related to your general result for $\delta I$ in part (a)?